$$\eqalign{ & {\text{We have }}g\left( { - 3} \right) = 0 \cr & \Rightarrow f\left( {g\left( { - 3} \right)} \right) = f\left( 0 \right) = 7{\left( 0 \right)^2} + 0 - 8 = - 8 \cr & \therefore \,{\text{fog}}\,\left( { - 3} \right) = - 8 \cr & g\left( 9 \right) = {9^2} + 4 = 85 \cr & \Rightarrow f\left( {g\left( 9 \right)} \right) = f\left( {85} \right) = 8.85 + 3 = 683 \cr & \therefore \,{\text{fog}}\,\left( 9 \right) = 683 \cr & f\left( 0 \right) = {7.0^2} + 0 - 8 = - 8 \cr & \Rightarrow g\left( {f\left( 0 \right)} \right) = g\left( { - 8} \right) = \left| { - 8} \right| = 8 \cr & f\left( 6 \right) = 4.6 + 5 = 29 \cr & \Rightarrow g\left( {f\left( 6 \right)} \right) = g\left( {29} \right) = {\left( {29} \right)^2} + 4 = 845 \cr & \therefore \,{\text{gof}}\,\left( 6 \right) = 845 \cr} $$

$$\eqalign{ & {\text{Let }}{x_1},\,{x_2}\, \in \,A{\text{ and let }}f\left( {{x_1}} \right) = f\left( {{x_2}} \right) \cr & {\text{or }}\frac{{{x_1} - 2}}{{{x_1} - 3}} = \frac{{{x_2} - 2}}{{{x_2} - 3}} \cr & {\text{or }}{x_1} = {x_2} \cr} $$
So, $$f$$ is one-one.
To find whether $$f$$ is onto or not, first let us find the range of $$f.$$
Let $$y = f\left( x \right) = \frac{{x - 2}}{{x - 3}}{\text{ or }}x = \frac{{3y - 2}}{{y - 1}}$$
$$x$$ is defined if $$y \ne 1,$$ i.e., the range of $$f$$ is $$R - \left\{ 1 \right\}$$  which is also the co-domain of $$f.$$
Also, for no value of $$y,\,x$$  can be $$3,$$ i.e., if we put $$3 = x = \frac{{3y - 2}}{{y - 1}}$$    then $$3y - 3 = 3y - 2{\text{ or }} - 3 = - 2$$       which is not possible. Hence, $$f$$ is onto.

Graphically, $$y = f\left( x \right) = \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right),$$       which is clearly many-one and onto.

Let $$n\left( P \right) = $$   Number of teachers in physics
$$n\left( M \right) = $$   Number of teachers in maths
$$\eqalign{ & n\left( {P \cup M} \right) = n\left( P \right) + n\left( M \right) - n\left( {P \cap M} \right) \cr & \Rightarrow \,20 = n\left( P \right) + 12 - 4 \cr & \Rightarrow \,n\left( P \right) = 12 \cr} $$
Only physics teachers $$= 8.$$

The smallest equivalence relation is the identity relation $${R_1} = \left\{ {\left( {a,\,a} \right),\,\left( {b,\,b} \right),\,\left( {c,\,c} \right)} \right\}$$
Then two ordered pairs of two distinct elements can be added to give three more equivalence relations $${R_2} = \left\{ {\left( {a,\,a} \right),\,\left( {b,\,b} \right),\,\left( {c,\,c} \right),\,\left( {a,\,b} \right),\,\left( {b,\,a} \right)} \right\}$$
Similarly $${R_3}$$ and $${R_4}.$$ Finally the largest equivalence relation i.e., the universal relation.

Numbers which are divisible by $$5$$ are $$5,\,10,\,15,\,20,\,25,\,30,\,35,\,40,\,45,\,50,\,55,\,60,\,65,\,70,\,75,\,80$$            they are $$16$$  in numbers. Now, numbers which are divisible by $$7$$ are $$7,\,14,\,21,\,28,\,35,\,42,\,49,\,56,\,63,\,70,\,77$$        they are $$11$$  in numbers.
Also, total odd numbers $$=40$$
Let $$C$$ represents the students who opt. for Cricket, $$F$$ for Football and $$H$$ for Hockey.
$$\therefore \,{\text{we have }}n\left( C \right) = 40,\,\,n\left( F \right) = 16,\,\,n\left( H \right) = 11$$
Now, $$C \cap F = $$   Odd numbers which are divisible by $$5.$$
$$C \cap H = $$   Odd numbers which are divisible by $$7.$$
$$F \cap H = $$   Numbers which are divisible by both $$5$$ and $$7.$$
$$\eqalign{ & n\left( {C \cap F} \right) = \,8,\,\,n\left( {C \cap H} \right) = 6,\,\,n\left( {F \cap H} \right) = 2,\,\,n\left( {C \cap F \cap H} \right) = 1 \cr & {\text{we know}} \cr & n\left( {C \cup F \cup H} \right) = n\left( C \right) + n\left( F \right) + n\left( H \right) - n\left( {C \cap F} \right) - n\left( {C \cap H} \right) - n\left( {F \cap H} \right) + n\left( {C \cap F \cap H} \right) \cr & {\text{or, }}n\left( {C \cup F \cup H} \right) = 67 - 16 + 1 = 52 \cr & \therefore \,n\left( {C' \cap F' \cap H'} \right) = {\text{Total students}} - n\left( {C \cup F \cup H} \right) \cr & {\text{or, }}n\left( {C' \cap F' \cap H'} \right) = 80 - 52 = 28 \cr} $$

Since $$\sin \left( {2x - 3} \right)$$   is a periodic function with period $$\pi ,$$ therefore $$f$$ is not injective. Also, $$f$$ is not surjective as its range $$\left[ { - 1,\,1} \right]$$  is a proper subset of its co-domain $$R.$$

From Venn-Euler's Diagram

$$\therefore \,\left( {A \cup B} \right)' \cup \left( {A' \cap B} \right) = A'$$

$$R$$ is defined over the set of non-negative integers, $${x^2} + {y^2} = 36$$
$$ \Rightarrow y = \sqrt {36 - {x^2}} = \sqrt {\left( {6 - x} \right)\left( {6 + x} \right)} ,\,x = 0{\text{ or }}6$$
for $$x = 0,\,y = 6$$   and for $$x = 6,\,y = 0$$
So, $$y$$ is $$6$$ or $$0$$
So, $$R = \left\{ {\left( {6,\,0} \right),\,\left( {0,\,6} \right)} \right\}$$

Reflexive and transitive only.
e.g. (3, 3), (6, 6), (9, 9), (12, 12) [Reflexive]
(3, 6), (6, 12), (3, 12) [Transitive].
(3, 6) $$ \in R$$  but (6,3) $$ \notin R$$  [ non symmetric]